$\int t^8\,dt=$ $+C$
Answer: The integrand is of the form $x^n$ where $n\neq-1$, so we can use the reverse power rule: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ $\begin{aligned} \int t^{{8}}\,dt&=\dfrac{t^{{8}+1}}{{8}+1}+C \\\\ &=\dfrac19 t^9+C \end{aligned}$ In conclusion, $\int t^8\,dt=\dfrac19 t^9+C$